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1/5.6 + 1/6.7 + 1/7.8 +...+ 1/24.25
=1/5 - 1/6 + 1/6-1/7 +1/7-1/8 + ... + 1/24-1/25
=> Kết quả là: 1/5 - 1/25 = 4/25
b) 2/1.3 + 2/3.5 + 2/5.7 + 2/7.9+...+ 2/99.101
=2/1-2/3 + 2/3-2/5 + 2/5-2/7 + 2/7-2/9 + ... + 2/99-2/101
=> kết quả là 2/1 - 2/101 =200/101
a) \(\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{24.25}\)
=\(\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{24}-\frac{1}{25}\)
=\(\frac{1}{5}-\frac{1}{25}\)
=\(\frac{4}{25}\)
b)\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{99.101}\)
=\(2.\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{99.101}\right)\)
=\(2.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\right)\)
=\(2.\left(\frac{1}{1}-\frac{1}{101}\right)\)
=\(2.\frac{100}{101}\)
=\(\frac{200}{101}\)
\(S1=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+....+\frac{2}{99.101}\)
\(S1=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-....-\frac{1}{101}=\frac{1}{1}-\frac{1}{101}=\frac{100}{101}\)
\(S2=\frac{5}{1.3}+\frac{5}{3.5}+....+\frac{5}{99.101}\)
\(S2=\frac{5}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-.....-\frac{1}{101}\right)=\frac{5}{2}.\left(\frac{1}{1}-\frac{1}{101}\right)=\frac{5}{2}\cdot\frac{100}{101}=\frac{250}{101}\)
a, \(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{99.101}\)
=2.(\(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{99.101}\))
=\(2.\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\right)\)
=\(\frac{2}{2}.\left(1-\frac{1}{101}\right)\)
\(=\frac{100}{101}\)
b, \(\frac{5}{1.3}+\frac{5}{3.5}+...+\frac{5}{99.101}\)
=\(5.\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{99.101}\right)\)
=\(5.\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(=\frac{5}{2}.\left(1-\frac{1}{101}\right)\)
=\(\frac{250}{101}\)
\(=\frac{5}{2}.\frac{100}{101}\)
a,21.321.3+23.523.5+25.725.7+....+299.101
=>\(\frac{1}{1}-\frac{1}{3}+\frac{1}{5}-\frac{1}{5}+....+\frac{1}{99}-\frac{1}{101}\)
=>\(\frac{1}{1}-\frac{1}{101}\)
=>\(\frac{100}{101}\)
b,
51.351.3+53.553.5+55.755.7+....+599.101
=>\(\frac{5}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+....+\frac{2}{99.101}\right)\)
=>\(\frac{5}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{99}-\frac{1}{101}\right)\)
=>\(\frac{5}{2}\left(\frac{1}{1}-\frac{1}{101}\right)\)
=>\(\frac{5}{2}.\frac{100}{101}\)
=>\(\frac{250}{101}\)
=1-1/3+1/3-1/5+1/5-1/7+...+1/99-1/101
=1-1/101
=100/101
k cho mình nha
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\)
\(=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{99}-\frac{1}{101}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{101}\right)=\frac{1}{2}.\frac{100}{101}=\frac{50}{101}\)
A = 1/1.3 + 1/3.5 + 1/5.7 +........+ 1/1999.2001
2.A = 2/1.3 + 2/3.5 + 2/5.7 +........+ 2/1999.2001
2.A = 1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + ..... + 1/1999 - 1/2001
2.A = 1 - 1/2001
2.A = 2000/2001
Vậy A =1000/2001
B = 1/3.5 + 1/5.7 + 1/7.9 +........+ 1/99.101
2.A = 2/3.5 + 2/5.7 + 2/7.9 +........+ 2/99.101
2.A = 1/3 - 1/5 + 1/5 - 1/7 + ..... + 1/99 - 1/101
2.A = 1/3 - 1/101 = 98/303
Vậy A =49/303
\(A=\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{1999.2001}\)
\(2A=\frac{2}{1.3}+\frac{2}{3.5}+....+\frac{2}{1999.2001}\)
\(2A=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{1999}-\frac{1}{2001}\)
\(2A=\frac{1}{1}-\frac{1}{2001}=\frac{2000}{2001}\)
\(A=\frac{2000}{2001}.\frac{1}{2}=\frac{1000}{2001}\)
= 2 x [1 - 1/3 + 1/3 - 1/5 + 1/5 -1/7 +1/7 -1/9 + .., +1/99 - 1/101
= 2 x [ 1 - 1/101 ]
= 2 x 100/101
= 200/101
t cho mik nha
\(\frac{2}{1.3}\)+\(\frac{2}{3.5}\)+\(\frac{2}{5.7}\)+.........+\(\frac{2}{99.101}\)
=\(\frac{1}{1}\)-\(\frac{1}{3}\)+\(\frac{1}{3}\)-\(\frac{1}{5}\)+\(\frac{1}{5}\)-\(\frac{1}{7}\)+....+\(\frac{1}{99}\)-\(\frac{1}{101}\)
= 1 - \(\frac{1}{101}\)= \(\frac{100}{101}\)
Bạn gõ lại đề đi :v
Đọc chả hiểu đề gì cả ... đề k có x
Mà phía dưới có cái đáp số x= ... là sao ??
a)(\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{11.12}\)). x=\(\frac{1}{3}\)
(1-\(\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...-\frac{1}{11}_{ }+\frac{1}{12}\)).x=\(\frac{1}{3}\)
(1+\(\frac{1}{12}\)).x=\(\frac{1}{3}\)
x=\(\frac{1}{3}:\frac{13}{12}\)
x=\(\frac{4}{13}\)
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{99.101}\)
Đặt A = \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{99.101}\)
\(2A=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\)
\(2A=\frac{1}{1}-\frac{1}{101}\)
\(2A=\frac{100}{101}\)
\(\Rightarrow A=\frac{100}{101}\div2\)
\(\Rightarrow A=\frac{50}{101}\)
đề
sai r bn ak